differentiate. y = ln(x + 7)^3(x + 8)^2(x + 1)^4 d/dxln(x + 7)^3(x + 8)^2(x + 1)^4 = □

Explanation:

Step1: Use log - property

First, use the property $\ln(abc)=\ln a+\ln b+\ln c$. So, $y = \ln[(x + 7)^{3}(x + 8)^{2}(x + 1)^{4}]=\ 3\ln(x + 7)+2\ln(x + 8)+4\ln(x + 1)$.

Step2: Differentiate term - by - term

The derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$ by the chain - rule. For $u=x + 7$, the derivative of $3\ln(x + 7)$ is $3\times\frac{1}{x + 7}\times1=\frac{3}{x + 7}$. For $u=x + 8$, the derivative of $2\ln(x + 8)$ is $2\times\frac{1}{x + 8}\times1=\frac{2}{x + 8}$. For $u=x + 1$, the derivative of $4\ln(x + 1)$ is $4\times\frac{1}{x + 1}\times1=\frac{4}{x + 1}$.

Step3: Sum up the derivatives

The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}=\frac{3}{x + 7}+\frac{2}{x + 8}+\frac{4}{x + 1}$.

Answer:

$\frac{3}{x + 7}+\frac{2}{x + 8}+\frac{4}{x + 1}$